Question 3 Let\'s say we wanted to analyze a model of alcohol consumption. The m

Question

Question 3 Let's say we wanted to analyze a model of alcohol consumption. The model looks at whether education, doctor visits, marital status, and employment status predict alcohol consumption. We have 10,000 observations. For hypothesis tests use the standard normal distribution. The equation we estimated is below. Standard errors are in parentheses 0 Di = 13.00 + 11 .36DOG-. 20 EDUCit 2.85DIVt 14.20UNEMP (7.12) (2.12) (.31) (2.55) where: D-drinks consumed by the ith individual in the last two weeks DOC 1 if the person has gone to the doctor in the last year, 0 if not EDUC years of schooling of the ith individual DIV 1 if the ith individual is divorced, 0 if not UNEMP-1 if the ith individual is unemployed, 0 if not It seems reasonable to expect positive coefficients for DIV and UNEMP. Conduct a test at the 5% level of the null hypothesis that the coefficient on DIV0?

Explanation / Answer

H0 : Coeff. of DIV =0

Ha : Coeff of DIV<=0

t = (2.85 - 0)/2.55 = 1.118

Z0.95 = -1.64

Hence, t > Z0.95. So, we cannot reject the null hypothesis

If we check DIV>0, the null and alternate hypothesis are:

H0 : Coeff. of DIV = 0

Ha : Coeff of DIV>0

t = (2.85 - 0)/2.55 = 1.118

Z0.05 = 1.64.

Hence, t < Z0.05. So we cannot reject the null hypothesis. So, our conclusion does not change

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